The matter will be consumed by other living organisms and the blood will settle to the bottom of the body
Answer:
So, you're dealing with a sample of cobalt-60. You know that cobalt-60 has a nuclear half-life of
5.30
years, and are interested in finding how many grams of the sample would remain after
1.00
year and
10.0
years, respectively.
A radioactive isotope's half-life tells you how much time is needed for an initial sample to be halved.
If you start with an initial sample
A
0
, then you can say that you will be left with
A
0
2
→
after one half-life passes;
A
0
2
⋅
1
2
=
A
0
4
→
after two half-lives pass;
A
0
4
⋅
1
2
=
A
0
8
→
after three half-lives pass;
A
0
8
⋅
1
2
=
A
0
16
→
after four half-lives pass;
⋮
Explanation:
now i know the answer
The correct answers are
Free enterprise and government
:)
Answer:
[Ag⁺] = 0.0666M
Explanation:
For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:
Ag⁺ + 2CN⁻ ⇄ Ag(CN)₂⁻
Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]
As initial concentrations of Ag⁺ and CN⁻ are:
[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M
[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M
The equilibrium concentrations of each compound are:
[CN⁻] = 9.7x10⁻⁴M - x
[Ag⁺] = 0.0676M - x
[Ag(CN)₂⁻] = x
<em>Where x is reaction coordinate</em>
Replacing in Kf formula:
1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]
1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0
Solving for x:
X = 9.614x10⁻⁴M
Thus, equilibrium concentration of Ag⁺ is:
[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>
